Question

Two particles, an electron and a proton, are initially at rest in a uniform electric field of magnitude 478 N/C. If the particles are free to move, what are their speeds (in m/s) after 52.0 ns?

Give the speed of both the electron and proton.

Answer 1

For electron,

E=478N/C, q=-1.6*10^-19 C, v_i = 0 m/s, t=52.0*10^-9 s

F=qE = (-1.6*10^-19C)*(478N/C) = -7.65*10^-17 N

a=F/m = (-7.65*10^-17 N)/(9.1*10^-31kg) = -8.4*10^13 m/s^2

v_f = v_i +at

Plug values,

v_f = 0+(-8.4*10^13)*(52.0*10^-9)

v_f = -4.37*10^6 m/s…….directed opposite the direction electric field E

For proton,

E=478N/C, q=1.6*10^-19 C, v_i = 0 m/s, t=52.0*10^-9 s

F=qE = (1.6*10^-19C)*(478N/C) = 7.65*10^-17 N

a=F/m = (7.65*10^-17 N)/(1.67*10^-27kg) = 4.58*10^10 m/s^2

v_f = v_i +at

Plug values,

v_f = 0+(4.58*10^10)*( 52.0*10^-9)

v_f = 2381.6 m/s…….directed in the direction electric field E